Calorimetry mixture problem

In summary, the conversation discusses a problem involving the mixing of hot lead and warm water in a calorimeter and determining the values of q(lead), q(water), and q(cal). The specific heat of water and lead are provided and the final temperature of the mixture is given. The question then arises regarding the calculation of q(lead) and why 100 degrees is subtracted from 28.8 degrees. It is clarified that this subtraction is based on the initial temperatures and the negative value for q(lead) makes sense as it indicates that lead lost heat during the mixing process.
  • #1
Bashyboy
1,421
5
Hello, my problem is as follows: " If 150 g of lead at 100°C were placed in a calorimeter with 50 g of water at 28.8°C and the resulting temperature of the mixture was 22°C, what are the values of q(lead), q(water), and q(cal)? (Knowing that the specific heat of water is 4.184 J/g °C and the specific heat of lead is 0.128 J/g °C)

The part that I don't understand in the calculations is this:
q(lead) = 0.128 J/g °C x 150g x (28.8°C - 100°C) = -1.37 E3 J

Why do they subtract 100 from 28.8? From my present understanding, 100 degrees is the initial temperature of the lead and 28.8 degrees is the initial temperature of water; shouldn't they subtract 100 degrees from 22 degrees?

Thank you.
 
Chemistry news on Phys.org
  • #2
No way putting hot lead in a warm water can yield a mix colder than initial temperatures of both substances.

q is mcΔT, where ΔT=Tfinal-T[/sub]initial[/sub], so at least the second part (-100) is correct. It yields a negative q, meaning lead lost heat - that also makes sense, and fits the convention.
 

1. What is a calorimetry mixture problem?

A calorimetry mixture problem is a type of problem commonly encountered in chemistry that involves calculating the final temperature of a mixture of two substances with different temperatures. This type of problem is often used to determine the specific heat capacity of a substance or to calculate the heat released or absorbed in a chemical reaction.

2. How do you solve a calorimetry mixture problem?

To solve a calorimetry mixture problem, you first need to calculate the heat released or absorbed by each substance using the formula Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Next, you can use the principle of conservation of energy to set the total heat released or absorbed by the substances equal to each other and solve for the final temperature of the mixture.

3. What is the principle of conservation of energy in calorimetry?

The principle of conservation of energy in calorimetry states that in a closed system, the total amount of energy remains constant. This means that the heat released or absorbed by one substance must be equal to the heat released or absorbed by the other substance in a calorimetry mixture problem.

4. What are the units of specific heat capacity?

The units of specific heat capacity are Joules per gram per degree Celsius (J/g°C) in the SI system or calories per gram per degree Celsius (cal/g°C) in the metric system.

5. What are some common sources of error in calorimetry mixture problems?

Some common sources of error in calorimetry mixture problems include heat loss to the surroundings, incomplete mixing of the substances, and inaccurate measurement of the initial and final temperatures. It is important to properly insulate the calorimeter and carefully measure the temperatures to reduce these sources of error.

Similar threads

  • Chemistry
Replies
4
Views
1K
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
889
Replies
1
Views
818
Replies
1
Views
2K
  • Biology and Chemistry Homework Help
Replies
8
Views
2K
  • Biology and Chemistry Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Biology and Chemistry Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
6K
Back
Top